b, Ta có : \(0\le x\le1\)

\(\Rightarrow-2\le x-2\le-1< 0\)

Ta có : \(y=f\left(x\right)=2\left(m-1\right)x+\dfrac{m\left(x-2\right)}{\left(2-x\right)}\)

\(=2\left(m-1\right)x-m< 0\)

TH1 : \(m=1\) \(\Leftrightarrow m>0\)

TH2 : \(m\ne1\) \(\Leftrightarrow x< \dfrac{m}{2\left(m-1\right)}\)

Mà \(0\le x\le1\)

\(\Rightarrow\dfrac{m}{2\left(m-1\right)}>1\)

\(\Leftrightarrow\dfrac{m-2\left(m-1\right)}{2\left(m-1\right)}>0\)

\(\Leftrightarrow\dfrac{2-m}{m-1}>0\)

\(\Leftrightarrow1< m< 2\)

Kết hợp TH1 => m > 0

Vậy ...
 

\(x^2-2\left(m-1\right)x-m^3+\left(m+1\right)^2=0\)

Để pt có hai nghiệm thỏa mãn

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\x_1+x_2=2\left(m-1\right)\le4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m-2\right)\left(m+2\right)\ge0\\m\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m\in\left[-2;0\right]\cup\left(2;+\infty\right)\cup\left\{2\right\}\\m\le3\end{matrix}\right.\)\(\Rightarrow m\in\left[-2;0\right]\cup\left[2;3\right]\)

\(P=x^3_1+x_2^3+x_1x_2\left(3x_1+3x_2+8\right)\)

\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_1\left(x_1+x_2\right)+8x_1x_2\)

\(=8\left(m-1\right)^3+8\left(-m^3+m^2+2m+1\right)\)

\(=-16m^2+40m\)

Vẽ BBT với \(f\left(m\right)=-16m^2+40m\) ;\(m\in\left[-2;0\right]\cup\left[2;3\right]\)

Tìm được \(f\left(m\right)_{min}=-144\Leftrightarrow m=-2\)

\(f\left(m\right)_{max}=16\Leftrightarrow m=2\)

\(\Rightarrow P_{max}=16;P_{min}=-144\)

Vậy....

Bài 4 :

24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ

Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0 

Suy ra quãng đường AB là 36x(km)

Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)

Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)

Ta có phương trình: 

\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)

Vậy quãng đường AB dài 36.2 = 72(km)

`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

`=>(x-3)(x+3)=0`

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

`=>(x+3)(-x)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Do có quá ít câu hỏi nên bạn nào trả lời được, mình sẽ xóa khỏi mục "Câu hỏi hay" nhé!

Quoc Tran Anh Le CTV Chưa ra bài tiếp à!?

Đặt: x² - 4x = a thì pt thành

a(a + 8) + 16 = 0

<=> a² + 8a + 16 = 0

<=> (a + 4)² = 0

<=> a = - 4

<=> x² - 4x + 4 = 0

<=> (x - 2)² = 0

<=> x = 2

alibaba nguyễn: bạn ơi cho mình hỏi xíu ạ: a= -4 thì x^2-4x+4 ở đâu ra ạ??

a: Đặt \(a=\frac{1}{x-y+1};b=\frac{1}{x+y-2}\)

Theo đề, ta có: \(\begin{cases}-3a+b=12\\ 2a-3b=-1\end{cases}\Rightarrow\begin{cases}-9a+3b=36\\ 2a-3b=-1\end{cases}\)

=>\(\begin{cases}-9a+3b+2a-3b=36-1\\ -3a+b=12\end{cases}\Rightarrow\begin{cases}-7a=35\\ b=3a+12\end{cases}\)

=>\(\begin{cases}a=-5\\ b=3\cdot\left(-5\right)+12=-15+12=-3\end{cases}\Rightarrow\begin{cases}x-y+1=-\frac15\\ x+y-2=-\frac13\end{cases}\)

=>\(\begin{cases}x-y=-\frac15-1=-\frac65\\ x+y=-\frac13+2=\frac53\end{cases}\Rightarrow\begin{cases}x=\left(-\frac65+\frac53\right):2=\left(-\frac{18}{15}+\frac{25}{15}\right):2=\frac{7}{15}:2=\frac{7}{30}\\ y=\frac53-\frac{7}{30}=\frac{50}{30}-\frac{7}{30}=\frac{43}{30}\end{cases}\)

b: \(\begin{cases}x^2+2\left(y^2+2y\right)=10\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\Rightarrow\begin{cases}3x^2+6\left(y^2+2y\right)=30\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\)

=>\(\begin{cases}3x^2+6\left(y^2+2y\right)-3x^2+\left(y^2+2y\right)=30-9\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\)

=>\(\begin{cases}7\left(y^2+2y\right)=21\\ 3x^2=\left(y^2+2y\right)+9\end{cases}\Rightarrow\begin{cases}y^2+2y=3\\ 3x^2=3+9=12\end{cases}\Rightarrow\begin{cases}y^2+2y-3=0\\ x^2=4\end{cases}\)

=>\(\begin{cases}\left(y+3\right)\left(y-1\right)=0\\ x\in\left\lbrace2;-2\right\rbrace\end{cases}\Rightarrow\begin{cases}y\in\left\lbrace-3;1\right\rbrace\\ x\in\left\lbrace2;-2\right\rbrace\end{cases}\)

c: ĐKXĐ; x>1; y>-2

\(\begin{cases}\frac{7}{\sqrt{x-1}}-\frac{5}{\sqrt{y+2}}=\frac92\\ \frac{3}{\sqrt{x-1}}+\frac{2}{\sqrt{y+2}}=4\end{cases}\Rightarrow\begin{cases}\frac{21}{\sqrt{x-1}}-\frac{15}{\sqrt{y+2}}=\frac92\cdot3=\frac{27}{2}\\ \frac{27}{\sqrt{x-1}}+\frac{18}{\sqrt{y+2}}=36\end{cases}\)

=>\(\begin{cases}\frac{21}{\sqrt{x-1}}-\frac{15}{\sqrt{y+2}}-\frac{21}{\sqrt{x-1}}-\frac{18}{\sqrt{y+2}}=\frac{27}{2}-36\\ \frac{7}{\sqrt{x-1}}-\frac{5}{\sqrt{y+2}}=\frac92\end{cases}\Rightarrow\begin{cases}-\frac{33}{\sqrt{y+2}}=\frac{27}{2}-\frac{72}{2}=\frac{-45}{2}\\ \frac{7}{\sqrt{x-1}}=\frac{5}{\sqrt{y+2}}+\frac92\end{cases}\)

=>\(\begin{cases}\sqrt{y+2}=33\cdot\frac{2}{45}=\frac{66}{45}=\frac{22}{15}\\ \frac{7}{\sqrt{x-1}}=5:\frac{22}{15}+\frac92=5\cdot\frac{15}{22}+\frac92=\frac{75}{22}+\frac{99}{22}=\frac{174}{22}=\frac{87}{11}\end{cases}\)

=>\(\begin{cases}y+2=\frac{484}{225}\\ x-1=\frac{5929}{7569}\end{cases}\Rightarrow\begin{cases}y=\frac{484}{225}-2=\frac{34}{225}\\ x=\frac{13498}{7569}\end{cases}\) (nhận)

1: \(\left(x-1\right)\left(x+5\right)\left(x^2+4x+8\right)+40=0\)

=>\(\left(x^2+4x-5\right)\left(x^2+4x+8\right)+40=0\)

=>\(\left(x^2+4x\right)^2+3\left(x^2+4x\right)-40+40=0\)

=>\(\left(x^2+4x\right)^2+3\left(x^2+4x\right)=0\)

=>\(\left(x^2+4x\right)\left(x^2+4x+3\right)=0\)

=>x(x+4)(x+1)(x+3)=0

=>x∈{0;-4;-1;-3}

2: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-15=0\)

=>\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-15=0\)

=>\(\left(x^2-5x+4\right)^2+2\left(x^2-5x+4\right)-15=0\)

=>\(\left(x^2-5x+4+5\right)\left(x^2-5x+4-3\right)=0\)

=>\(\left(x^2-5x+9\right)\left(x^2-5x+1\right)=0\)

=>\(x^2-5x+1=0\)

=>\(x^2-5x+\frac{25}{4}-\frac{21}{4}=0\)

=>\(\left(x-\frac52\right)^2=\frac{21}{4}\)

=>\(x-\frac52=\pm\frac{\sqrt{21}}{2}\)

=>\(x=\frac52\pm\frac{\sqrt{21}}{2}\)

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)